It is enough to prove that there exists a non-zero weight vector \(z\)  with weight \(\mathbf {\lambda }\)  and an integer \(p\)  with the property that
\(\mathfrak {L}_{S, n}(B)[{(p + j}) \mathbf {\beta }+ K] .z = 0\)  for all \(j \ge 1\) , where \(\mathbb {Z}^n = \mathbb {Z}\mathbf {\beta }\oplus K\) . Then by irreducibilty of \(V\)  and PBW theorem, it will imply that the \(\mathrm {Supp}(V) \subseteq \lbrace \mathbf {\lambda }+ (p - j)\mathbf {\beta }+ K: j \in \mathbb {Z}_{+} \rbrace \) . Hence we will have
\(V = U(\mathfrak {L}_{S, n} (B)).Z\) , where \(Z = U(\mathfrak {L}_{S, n} (B) [K]).z = V[{\mathbf {\lambda }+ p \mathbf {\beta }+ K}]\)  is the highest weight space of \(V\) .
Now, since \(V\)  is irreducible, there exists \({\mathbf {\mu }} \in n\)  such that Supp \(V \subseteq {\mathbf {\mu }} + \mathbb {Z}^n\) . Choose a \({\mathbf {\gamma }}\in n\)  with the property that \((\mathbf {{\mathbf {\gamma }}}|\bf {m}) \ne 0\)  for all \(\bf {m}\in \mathbb {Z}^n \setminus \lbrace 0\rbrace \) . Then consider \(V\)  as \(\mathrm {Vir}(\mathbf {\mathbf {\gamma }})\) -module and by exactly similar argument as that of Theorem 4.3 of [1]} we get \(\bf {m}\in \mathbb {Z}^n\)  such that \(m_1=0\)  and
\(\mathrm {dim} V_{{\mathbf {\mu }}-\bf {m}}>(n+ \mathrm {dim}\, S)(\mathrm {dim}V_{{{\mathbf {\mu }}+\bf {e}}_1}+\sum _{i=2}^{n}\mathrm {dim}V_{{\mathbf {\mu }}+{\bf {e}}_1+{\bf {e}}_i})\) 
Let \({\bf {q}}_1={\bf {m}}+{\bf {e}}_1\) , \({\bf {q}}_i={\bf {m}}+{\bf {e}}_1+{\bf {e}}_i\)  for \( i=2,\dots ,n\) . For simplification, let \(\lambda := {{\mathbf {\mu }}}-\bf {m}\) . Then by (4.3) we have non-zero vector \(v \in V_\lambda \)  such that \(D({\bf {e}}_i, {\bf {q}}_j). v= 0\)  and \(S\otimes t^{{\bf {q}}_j}.v=0\)  for all \(i,j=1,\dots ,n\) . It follows that \(v\)  is an \(\mathfrak {L}_{S,n}\) -generalised highest weight vector. Now consider the \(\mathfrak {L}_{S,n}\) -submodule \(W:=U(\mathfrak {L}_{S,n}).v\)  of \(V\) . Let \(\overset{\circ }{W}\)  be the submodule of \(W\)  containing trivial vectors We will prove at the end of this section (Proposition
REF ) that \(\overset{\circ }{W} = \lbrace {\bf 0}\rbrace .\), i.e., \(\overset{\circ }{W} = \lbrace v \in W \mid \mathfrak {L}_{S,n} . v = 0\rbrace \) .
Consider the quotient module \(\overline{W}:= W/ \overset{\circ }{W}\) .
Claim 1: An \(\mathfrak {L}_{S,n}\) -module \(\overline{W}\)  contains no non-zero trivial vectors.
Suppose contrary, there exists a trivial vector \(\bar{w} = w + \overset{\circ }{W}\) , with \(w \notin \overset{\circ }{W}\)  such that \(\mathfrak {L}_{S,n}w \in \overset{\circ }{W}\) . But as \(\bar{w}\)  is a trivial vector, it must lie in the zero weight space of \(\overline{W}\) . It implies that the weight of \(w\)  is also a zero
weight vector considered as a weight vector of \(W\) .
But we have \((\mathfrak {L}_{S,n})_{\pm {\bf e_i}}.w \in \overset{\circ }{W}\) , and as all vectors \(\overset{\circ }{W}\)  have weight zero (considered as a weight vectors of W), it follows that \((\mathfrak {L}_{S,n})_{\pm {\bf e_i}}. w = 0 \) . Hence we have \(\mathfrak {L}_{S,n}.w = 0\) , which is absurd. This proves claim 1. Let \(\bar{v}\)  denote the image of \(v\)  under the natural map from \(W\)  to \(\overline{W}\) . Then we have \(\overline{W} =U( \mathfrak {L}_{S,n}). (\bar{v})\) .
Claim 2: \(\overline{W}\)  contains a highest weight \(\mathfrak {L}_{S,n}\) -module.
By claim 1, \(\overline{W}\)  contains no non-zero trivial vectors. We have the following:

Every vector in \(\overline{W}\)  is a generalised highest weight vector.

For any \(\mathbf {\alpha }\in \mathbb {N}^n\)  and any \(0\ne \bar{w} \in \overline{W}\) , \( (\mathfrak {L}_{S,n})_{-\mathbf {\alpha }}.\bar{w} \ne 0\) .

For \(\eta \in \)  Supp\((\overline{W})\) , and any \(\mathbf {\alpha }\in \mathbb {N}^n\) , the set \(\lbrace \ell \in \mathbb {Z}| \eta +\ell \mathbf {\alpha }\in \)  Supp\((\overline{W})\rbrace =(-\infty , s]\)  for some \(s\in \mathbb {Z}_{+}\) .

Upto a change of variables we have \(\lambda +\mathbf {\alpha }\notin \)  Supp\((\overline{W})\)  for any non-zero \(\mathbf {\alpha }\in \mathbb {Z}_{+}^n\) .

Proof of above statements are exactly similar to Lemma 3.3, 3.4, 3.5, 3.6 of [2]} respectively. By (d) it follows that \(\overline{W}\)  is not a uniformly bounded \(\mathfrak {L}_{S,n}\) -module. So let \(\mathfrak {L}_{S,n}(K,\mathbf {\beta },X)\)  be its irreducible sub-quotient. Now as every element of \(\overline{W}\)  is a generalised highest weight vector, replacing \(\bar{v}\)  by a suitable vector if necessary we can assume that
\(\mathbf {\lambda }-\mathbb {Z}_{+}\mathbf {\beta }+K \setminus \lbrace {\bf 0}\rbrace  \subseteq \)  Supp\((\overline{W})\) .
We also have the following properties (see Lemma 3.7, [2]})

\(\mathbf {\lambda }+\mathbf {\alpha }+K \nsubseteq \)  Supp(\(\overline{W}\) ) for all \({\mathbf {\alpha }} \in \mathbb {Z}_{+}^n \setminus \lbrace {\bf 0}\rbrace \) .

\(\exists ~ {\mathbf {\nu }}\in \mathbb {N}^n\)  such that \(K:= \lbrace {\bf {y}}\in \mathbb {Z}^n : ({\mathbf {\nu }|\bf {y})}= 0\rbrace \) .

\(\mathbf {\lambda }+\bf {y}\in \)  Supp(\(\overline{W}\) ) \(\forall ~ {\bf {y}}\in \mathbb {Z}^n\)  with \((\mathbf {\nu }|\bf {y})<0\) .

Subclaim:
\(\lbrace \mathbf {\lambda }+k\mathbf {\beta }+ K: k\in \mathbb {N}\rbrace \bigcap \)  Supp\((\overline{W})\ne \phi \)  for only finitely many \(k\in \mathbb {N}\) .
Assume contrary, so we get a strictly increasing sequence of \(k_i\in \mathbb {N}\)  such that
\(\lbrace \mathbf {\lambda }+k_i\mathbf {\beta }+K: i\in \mathbb {N}\rbrace \bigcap \)  Supp\((\overline{W})\ne \phi \) .
Now by choosing sufficiently large \(k_i\in \mathbb {N}\)  and taking a subsequence of \(\lbrace k_i\in \mathbb {N}\rbrace \)  if necessary we can write
\( |\lbrace \mathbf {\lambda }+k_r\mathbf {\beta }+K\rbrace \bigcap \lbrace \mathbf {\lambda }+\mathbb {Z}_{+}^n\rbrace |>1\) 
\(|\lbrace \mathbf {\lambda }+k_{r-1}\mathbf {\beta }+K\rbrace \bigcap \lbrace \mathbf {\lambda }+\mathbb {Z}_{+}^n\rbrace |>0\) 
By equation REF , we have that \(W[\mathbf {\lambda }+k_{r}\mathbf {\beta }+K]\)  is a non-uniformly bounded \(\mathfrak {L}_{S,n}[K]\) -module. Now following the argument of the proof of Lemma 3.7 line by line we arrive at a contradiction. This completes the proof of subclaim.
Let \(j_0\in \mathbb {Z}_{+}\)  be the largest integer such that
\( \lbrace \mathbf {\lambda }+j_0\mathbf {\beta }+K \rbrace \bigcap \mathrm {Supp}(\overline{W})\ne \phi .\) 
Consider \(\overline{W}[\mathbf {\lambda }+j_0\mathbf {\beta }+K]\)  as \(\mathfrak {L}_{S,n}[K]\) -module. Observe that \(\overline{W}[\mathbf {\lambda }+j_0\mathbf {\beta }+K]\)  is a uniformly bounded \(\mathfrak {L}_{S,n}[K]\) -module. Otherwise it would contain an irreducible \(\mathfrak {L}_{S,n}[K]\bigcap \mathrm {Vir}(\mathbf {\mathbf {\gamma }})\)  sub-quotient say \(\bar{Y}\)  which is not uniformly bounded. But then the unique irreducible quotient of
\(\mathrm {Ind}_{U((\mathfrak {L}_{S,n}[K] \cap \mathrm {Vir}({\mathbf {\gamma }}))+({\mathfrak {L}_{S,n}}^+ \bigcap \mathrm {Vir}(\mathbf {\mathbf {\gamma }})))}^{U(\mathrm {Vir}(\mathbf {\mathbf {\gamma }}))}\bar{Y}\) 
is not a Harish-Chandra module which will imply that \(\overline{W}\)  is not a Harish-Chandra module.
Now as \(\overline{W}[\mathbf {\lambda }+j_0\mathbf {\beta }+K]\)  is uniformly bounded \(\mathfrak {L}_{S,n}[K]\) -module, let \(\overline{Z}\)  be its non-zero irreducible submodule. It is easy to see that the module \(\bar{M} := U(\mathfrak {L}_{S,n}).\overline{Z}\)  is the required highest weight \(\mathfrak {L}_{S,n}\) -module
contained in \(\overline{W}\) . This completes the proof of claim 2. Now we need to deal with two cases:
Case 1: \({\bf 0} \notin \mathrm {Supp}({\overline{W}})\) . Note that in this case all the zero weight vectors of \(W\)  lie in \(\overset{\circ }{W}\) . Let \(\overline{Z}\)  be its non-trivial irreducible submodule of \(\overline{W}[\mathbf {\lambda }+j_0\mathbf {\beta }+K]\) .
Let \(0 \ne \bar{z} \in \overline{Z}\)  be any weight vector. Let \(z\)  be any preimage of \(\bar{z}\)  under the surjective \(\mathfrak {L}_{S,n}[K]\) -map \(\Phi : W[{\mathbf {\lambda }}+j_0{\mathbf {\beta }}+K] \rightarrow \overline{W}[{\mathbf {\lambda }}+j_0{\mathbf {\beta }}+K]\) . Let us define \(M := U(\mathfrak {L}_{S,n}).z\) . Let \(M^{\prime }\)  be the maximal
\(\mathfrak {L}_{S,n}\) -submodule of \(M\)  with property that \(M_{{\mathbf {\gamma }}}^{\prime } = 0\)  for all \(\mathbf {\gamma }\in \mathrm {Supp}(\overline{Z})\) . The existence of \(M^{\prime }\)  follows from the fact that \(M\)  is a cyclic module and equation REF . Observe also that \(M \cap \overset{\circ }{W} \subseteq M^{\prime }\) .
Let us consider the restriction of map \(\Phi \)  on an \(\mathfrak {L}_{S,n}[K]\) -module \(M \lbrace K\rbrace  := U(\mathfrak {L}_{S,n} [K]).z\) , which we denote by \(\psi \) . Then \(\psi : M \lbrace K\rbrace  \mapsto \overline{Z}\)  is a surjective \(\mathfrak {L}_{S,n} [K]\) -module map defined by \(\psi (u.z) = u.\bar{z}\) , where \(u \in \mathfrak {L}_{S,n} [K]\) . We have
\(\mathrm {Kernel}(\psi ) = M\lbrace K\rbrace  \cap \overset{\circ }{W}\) .
Claim 3: \(M/M^{\prime }\)  is an irreducible \(\mathfrak {L}_{S,n}\) -module.
Assume on the contrary that there exists a proper non-zero \(\mathfrak {L}_{S, n}\) -submodule \(M_1\)  of \(M\)  containing \(M^{\prime }\)  such that \((M_1)_{{\mathbf {\gamma }}} \ne 0\)  where \({\mathbf {\gamma }} \in \mathrm {Supp} (\overline{Z})\) .
Let \(0 \ne z_1 \in (M_1)_{{\mathbf {\gamma }}}\) . Then there exists \(u \in U(\mathfrak {L}_{S,n})\)  such that \(u.z = z_1\) .
Subclaim: \(u \in U(\mathfrak {L}_{S,n}[K])\) .
Let us assume that the subclaim is not true. Then \(u\)  must be of the form \(u_1 u_2 u_3\) , where \(u_1 \in U_0(\mathfrak {L}_{S,n}^-)\) , \(u_2 \in U(\mathfrak {L}_{S,n}[K])\)  and \(u_3 \in U_0(\mathfrak {L}_{S,n}^+)\) . Here, for any Lie algebra \(\mathfrak {g}\) , \(U_0(\mathfrak {g})\)  denotes the argumentation ideal of
\(U(\mathfrak {g})\) . But by equation REF , it follows that \(u_3.z = 0\)  unless weight of the vector \(u_3.z\)  is zero. But as all the zero weight vectors in M lies in \(\overset{\circ }{W}\) , it follows that \(u_3 .z \in \overset{\circ }{W}\)  and hence \(z_1 = 0\)  which is a required contradiction.
This proves the subclaim. Now, as \(\psi (z_1) = \bar{z_1}\)  is a non-zero element of \(\overline{Z}\)  and \(\overline{Z}\)  is an irreducible \(\mathfrak {L}_{S,n}[K]\) -module, there exists an element \(u_1 \in U(\mathfrak {L}_{S,n}[K])\)  such that \(u_1 . \bar{z_1} = \bar{z}\) . But this implies that \(z - u_1.z_1 \in M \cap \overset{\circ }{W}\subseteq M^{\prime }\) . So, we have \(z + M^{\prime } = u_1.z_1 + M^{\prime }\) , which is a contradiction to the assumption that \(M_1/M^{\prime }\)  is a proper submodule of \(M/M^{\prime }\) . This completes the proof of claim 3.
By Theorem REF , we have \(M/M^{\prime } \cong L_{\mathfrak {L}_{S,n}}(K, {\mathbf {\beta }}, Z)\) ,
where \(Z = U({\mathfrak {L}_{S, n}} [K]).\bar{z}^{\prime }\) , where \(\bar{z}^{\prime }\)  is image of \(z\)  under the natural map from \(M\)  to \(M/M^{\prime }\) .
Let \({\bf {g}}_2\in K\)  such that \(V_{\mathbf {\lambda }+j_0 {\mathbf {\beta }}+{\bf {g}}_2}\ne 0\) . As \(L_{\mathfrak {L}_{S,n}}(K,{\mathbf {\beta }}, Z)\)  is non-uniformly bounded there exists \({\bf {g}}_1\in K\)  and \(p\in \mathbb {N}\)  such that
\(\mathrm {dim}{(L_{\mathfrak {L}_{S,n}}(K,{\mathbf {\beta }}, Z))}_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1} )> (\mathrm {dim}\,S+n)\mathrm {dim} V_{\mathbf {\lambda }+j_0\mathbf {\beta }+{\bf {g}}_2}\) 
Let \(0\ne f\in B\) . Let \( M^{\prime \prime }_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1}\)  be a vector space compliment of \(M^{\prime }_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1}\)  in \(M_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1}\) . Then we have
\(\mathrm {dim}M^{\prime \prime }_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1}>(\mathrm {dim}\,\,S+n)\mathrm {dim}V_{\mathbf {\lambda }+j_0\mathbf {\beta }+{\bf {g}}_2}\) .
By above inequality there exists \(0\ne w_{f,p}^{{\bf {g}}_1}\in M_{\mathbf {\lambda }+(j_0-p)\mathbf {\beta }+{\bf {g}}_1}\)  such that
\({((\mathfrak {L}_{S,n})_{p\mathbf {\beta }+{\bf {g}}^{\prime }}\otimes f)w_{f,p}^{{\bf {g}}_1}=0}\) , where \( {\bf {g}}^{\prime } = {\bf g_2 - g_1}\) .
Let us choose \(N \in \mathbb {N}\)  be large enough so that \(\mathbf {\lambda }+ (p + j) \mathbf {\beta }+ K \ne \bf 0\)  for all \(j \ge N\) . Then we have
\({(\mathfrak {L}_{S, n})_{(p+j)\mathbf {\beta }+{\bf {g}}}.w_{f,p}^{{\bf {g}}_1}=0}\)  \(\forall j\ge N\)  and \(\forall {\bf {g}}\in K\) .
As \(L_{\mathfrak {L}_{S,n}}(K,{\mathbf {\beta }}, Z)\)  is irreducible, there exists \(u_1,\dots ,u_\ell \)  and \(h_1,\dots ,h_\ell \)  such that
\(h_1\cdots h_qu_1\cdots u_\ell .w_{f,p}^{{\bf {g}}_1} = \bar{z}^{\prime },\)
where
\(u_i\in (\mathfrak {L}_{S, n})_{(j_i){\mathbf {\beta }}+{\bf {g}}_i^{\prime \prime }}, 1\le i \le \ell , j_i>0, \sum _{i = 1}^{\ell }j_i = p, \,\, \mathrm {and}\,\, {\bf {g}}_i^{\prime \prime }\in K\,\, \mathrm {and}\,\, h_k\in (\mathfrak {L}_{S, n})_{{\bf {g}}_k^{\prime }},\)  \( {\bf g}_k^{\prime }\in K,\,\, \forall \,\, 1\le k\le q.\) 
But as \(M_{\mathbf {\gamma }}^{\prime } = \lbrace 0\rbrace \)  for all \({\mathbf {\gamma }} \in \mathrm {Supp} (\overline{Z})\) , we have \(h_1\cdots h_q u_1\cdots u_\ell .w_{f,p}^{{\bf {g}}_1} = z.\) 
Now, using
\({((\mathfrak {L}_{S, n})_{p\mathbf {\beta }+{\bf {g}}^{\prime }}\otimes f)w_{f,p}^{{\bf {g}}_1}=0} ={(\mathfrak {L}_{S, n})_{(p+j)\mathbf {\beta }+{\bf {g}}}.w_{f,p}^{{\bf {g}}_1}},\)  for all \(j \ge N\) ,
we have
\({((\mathfrak {L}_{S, n})_{(2p+j)\mathbf {\beta }+{\bf {g}}}\otimes f)w_{f,p}^{{\bf {g}}_1}} = [{(\mathfrak {L}_{S, n})_{(p+j)\mathbf {\beta }+{\bf {g-g}}_1}, {((\mathfrak {L}_{S, n})_{p\mathbf {\beta }+{\bf {g}}_1}\otimes f)}].w_{f,p}^{{\bf {g}}_1}}=0 \, \forall {\bf g} \in K.\) 
But then for all \(j\ge N\) ,
\(((\mathfrak {L}_{S, n})_{(2p+j)\mathbf {\beta }+{\bf {g}}}\otimes f)z=((\mathfrak {L}_{S, n})_{(2p+j)\mathbf {\beta }+{\bf {g}}}\otimes f)h_1\cdots h_qu_1\cdots u_\ell .w_{f,p}^{{\bf {g}}_1}=0\)  \(\forall {\bf g}\in K \)
by an induction on \(q + \ell \)  and using \(((\mathfrak {L}_{S, n})_{(2p+j)\mathbf {\beta }+{\bf {g}}}\otimes f).w_{f,p}^{{\bf {g}}_1}=0\)  \(\forall {\bf g}\in K, j\ge N.\)
 Case 2 : \({\bf 0} \in \mathrm {Supp}({\overline{W}})\) . In this case we have two subcases:
Subcase (i): \(\lbrace \mathbf {\lambda }+j_0\mathbf {\beta }+K\rbrace  \cap \mathrm {Supp} (\overline{W}) \ne \lbrace  \bf 0\rbrace \) .
Let \( \overset{\circ }{\overline{W}}[ \mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]\)  be a maximal trivial \(\mathfrak {L}_{S,n}[K]\) -module of
\(\overline{W}[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]\) . It follows that
\( \overset{\circ }{\overline{W}}[ \mathbf {\lambda }+ j_0 \mathbf {\beta }+ K] = \overset{\circ }{{W}}[K] \cap W[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]/ \overset{\circ }{W} \cap W[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K],\) 
where \(\overset{\circ }{W}[K] : = \lbrace w \in W : \mathfrak {L}_{S,n}[K].w = 0\rbrace \) . Now, \(\overline{W}[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]/ \overset{\circ }{\overline{W}}[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]\)  contains a non-trivial irreducible \(\mathfrak {L}_{S,n}[K]\) -submodule, which we denote by \(\overline{D}\) .
We have
\(W[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]/ \overset{\circ }{W}[K] \cap W [\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K] \cong \overline{W}[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]/ \overset{\circ }{\overline{W}}[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K] \) 
as \(\mathfrak {L}_{S,n}[K]\) -modules.
Consider a natural surjective \(\mathfrak {L}_{S,n}[K]\) -module map
\(\Phi : W[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K] \rightarrow W[\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]/\overset{\circ }{W}[K] \cap W [\mathbf {\lambda }+ j_0 \mathbf {\beta }+ K]. \) 
Let \(0 \ne \bar{s} \in \overline{D}\)  be a non-zero vector of non-zero weight. Let \(s\)  be any preimage of
\(\bar{s}\)  under \(\Phi \) . Let \(T = U(\mathfrak {L}_{S, n}).s\)  and \(T^{\prime }\)  be its maximal \(\mathfrak {L}_{S,n}\) -module with the property that \(T^{\prime }_{\mathbf {\eta }} = 0\)  for all \(\mathbf {\eta }\in \mathrm {Supp}(\overline{D}) \setminus \lbrace 0\rbrace \) . It is easy to see that \(T \cap \overset{\circ }{W}[K] \subseteq T^{\prime }\) . Let us define an \(\mathfrak {L}_{S,n}[K]\) -module \(T\lbrace K\rbrace : = U(\mathfrak {L}_{S,n}[K]).s\)  and an \(\mathfrak {L}_{S,n}[K]\) -module surjective map \(\psi : T\lbrace K\rbrace  \mapsto \overline{D}\)  by \(\psi (u.s) = u.\bar{s}\) , where \(u \in U(\mathfrak {L}_{S,n}[K]).\)  Note that \(\psi \)  is a restriction of the map \(\Phi \)  on \(T\lbrace K\rbrace \)  and we have
\(T \lbrace K\rbrace / T\lbrace K\rbrace \cap \overset{\circ }{W}[K] \cong \overline{D}.\)
Claim 4: \(T/T^{\prime }\)  is an irreducible \(\mathfrak {L}_{S,n}\) -module.
Let us assume that the claim is false. So there exist a proper \(\mathfrak {L}_{S,n}\) -submodule \(T_1\)  containing \(T^{\prime }\)  such that \((T_1)_{\mathbf {\gamma }} \ne 0\)  for some \(\mathbf {\gamma }\in \mathrm {Supp}(\overline{D}) \setminus \lbrace \bf 0\rbrace \) . Let \(0 \ne s_1 \in (T_1)_{\mathbf {\gamma }}.\)  Then there
exists \(u \in U(\mathfrak {L}_{S,n})\)  such that \(u.s = s_1\) . We see here that \(u\)  must lie in \(U(\mathfrak {L}_{S,n}[K])\) ; otherwise \(u = u_1 u_2 u_3\)  where \(u_1 \in U_0(\mathfrak {L}_{S,n}^-)\) , \(u_2 \in U(\mathfrak {L}_{S,n}[K])\)  and \(u_3 \in U_0(\mathfrak {L}_{S,n}^+)\) . But we notice that
\(u_3. s = 0\)  unless weight of \(u_3. s\)  is zero. But this possibility is ruled out by the choice of \(j_0\)  in the equation REF . Hence we have \(u \in U(\mathfrak {L}_{S,n}[K])\) . Now, consider the image \(\psi (s_1) = \bar{s_1}\) , which is non-zero as \(s_1 \notin T^{\prime }\) . But then irreducibility of
\(\overline{D}\)  implies that there exists an \(u_1 \in U(\mathfrak {L}_{S,n}[K])\)  such that \(u_1.(\bar{s_1}) = \bar{s}\) . But this forces that \(s + T^{\prime } = u_1.s_1 + T^{\prime }\) , which is a contradiction to the assumption that \(T_1/T^{\prime }\)  is a proper submodule of \(T/T^{\prime }\)  and we are done with the proof of claim 4.
Now, we have \(T/T^{\prime } \cong L_{\mathfrak {L}_{S,n}}(K, {\mathbf {\beta }}, D)\) , where \(D = U(\mathfrak {L}_{S,n}[K]) \bar{s}^{\prime }\) , where \( \bar{s}^{\prime }\)  is the image of \(s\)  in \(T/T^{\prime }\) . Now, rest of the proof follows exactly along the similar lines as that of proof of case 1.
Subcase (ii): \(\lbrace \mathbf {\lambda }+j_0\mathbf {\beta }+K\rbrace  \bigcap \mathrm {Supp} (\overline{W}) = \lbrace  \bf 0\rbrace \) . In this subcase notice that \(\overline{W}[\mathbf {\lambda }+j_0 \mathbf {\beta }+K] = \overset{\circ }{W}[K]/ \overset{\circ }{W} = \overline{W}_{\bf 0}\)  ,where
\(\overset{\circ }{W}[K] : = \lbrace w \in W : \mathfrak {L}_{S,n}[K].w = 0\rbrace \)  Now, let \(0 \ne \bar{w} = w + \overset{\circ }{W}\) . Let \(j_1\)  be the largest integer with the property that \(j_1 < j_0\)  and \(\lbrace \mathbf {\lambda }+j_1\mathbf {\beta }+K\rbrace  \bigcap \mathrm {Supp} (\overline{W}) \ne \phi \) .
But using the property of the Lie algebra \(\mathfrak {L}_{S,n}\)  that \([(\mathfrak {L}_{S,n})_{\bf m}, (\mathfrak {L}_{S,n})_{\bf s}] = (\mathfrak {L}_{S,n})_{\bf m + s}\)  for \({\bf m, s} \in \mathbb {Z}^n\)  and \({\bf m} \ne {\bf s}\) , it follows that \(j_1 = j_0 -1\) . For if \(\lbrace  \mathbf {\lambda }+ (j_0 - 1) \mathbf {\beta }+ K\rbrace  \bigcap \mathrm {Supp} (\overline{W}) = \phi \) ,
then it would imply \(\mathfrak {L}_{S,n}[- \mathbf {\beta }+ K].w = 0\) . But then using the before mentioned bracket property of \(\mathfrak {L}_{S,n}\) , it will follow that \(\mathfrak {L}_{S,n}^- .w = 0\) . As we already have \(\mathfrak {L}_{S,n}^+ .w = 0 = \mathfrak {L}_{S,n}[K].w\) , it follows that \(\mathfrak {L}_{S,n}.w = 0\) , which is a contradiction to the assumption
that \(w \notin \overset{\circ }{W}\) . Let \(x \in (\mathfrak {L}_{S,n})_{-\mathbf {\beta }+ k}\)  such that \(x.w \ne 0\) , for some \(k \in K\) . Now using the facts that \(\mathfrak {L}_{S,n}^+ .w = 0 = \mathfrak {L}_{S,n}[K].w\) , a quick calculation shows that \(\mathfrak {L}_{S,n}^+ . (x.w) = 0\) . Let \(z_1 := x.w\)  and consider the \(\mathfrak {L}_{S,n}[K]\) -module
\(Z_1 = U(\mathfrak {L}_{S,n}[K]).z_1\) . An easy check shows that \(\mathfrak {L}^+_{S,n}. y = 0\)  for all \(y \in Z_1\) . Observe that \(Z_1\)  must be a uniformly bounded \(\mathfrak {L}_{S,n}[K]\) -module. Otherwise, considering it as \(\mathrm {Vir}(\mathbf {\gamma }) \cap \mathfrak {L}_{S,n}[K]\) -module and taking its irreducible subquotient \(\overline{Z_1}\) , which has finite dimensional but non-uniformly bounded weight spaces, we get a unique irreducible quotient of \(\mathrm {Vir}(\mathbf {\gamma })\) -module
\(\mathrm {Ind}_{U((\mathfrak {L}_{S,n}[K] \cap \mathrm {Vir}({\mathbf {\gamma }}))+({\mathfrak {L}_{S,n}}^+ \bigcap \mathrm {Vir}(\mathbf {\mathbf {\gamma }})))}^{U(\mathrm {Vir}(\mathbf {\mathbf {\gamma }}))}\bar{Z_1},\) 
which does not have finite dimensional weight spaces. This will eventually imply that \(W\)  does not have finite dimensional weight spaces. Now, as \(Z_1\)  is uniformly bounded and \({\bf 0} \notin \mathrm {Supp}(Z_1)\) , it contains a non-trivial irreducible \(U(\mathfrak {L}_{S,n}[K])\) -submodule
say \(Z\) . Then \(M : = U(\mathfrak {L}_{S,n}).Z\)  is a highest weight \(\mathfrak {L}_{S,n}\) -module contained in \(W\) . Let \(M/M^{\prime }\)  be its unique non-trivial irreducible quotient. Now, we can proof that any \(0 \ne z \in Z\)  is an \(\mathfrak {L}_{S,n}(B)\)  highest weight vector by using the same argument as in case 1.
